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schepotkina [342]
3 years ago
14

A man invests $17,000 in two accounts.one account earns 5% interest per year,and the other earns 6.5%. If the total interest aft

er one year is $970,how much did he invest at each rate?
Mathematics
1 answer:
uranmaximum [27]3 years ago
4 0
Let x = the amount invested a 5% and 17,000 - x = the amount invested at 6.5%. Now you can write the equation: 
<span>0.05x + [0.065 (17,000 - x)] = 970 </span>
<span>Solve for x to get the amount invested at 5% and 17,000 - x will = the amount invested at 6.5% </span>
<span>x = 9,000 </span>
<span>17,000 - 9,000 = 8,000</span>
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For a certain​ drug, the rate of reaction in appropriate units is given by Upper R prime (t )equalsStartFraction 2 Over t plus 1
Tems11 [23]

Answer:

a) 8.13

b) 4.10

Step-by-step explanation:

Given the rate of reaction R'(t) = 2/t+1 + 1/√t+1

In order to get the total reaction R(t) to the drugs at this times, we need to first integrate the given function to get R(t)

On integrating R'(t)

∫ (2/t+1 + 1/√t+1)dt

In integration, k∫f'(x)/f(x) dx = 1/k ln(fx)+C where k is any constant.

∫ (2/t+1 + 1/√t+1)dt

= ∫ (2/t+1)dt+ ∫ (1/√t+1)dt

= 2∫ 1/t+1 dt +∫1/+(t+1)^1/2 dt

= 2ln(t+1) + 2(t+1)^1/2 + C

= 2ln(t+1) + 2√(t+1) + C

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When t = 1

R(1) = 2ln2 + 2√2

≈ 4.21

When t = 12

R(12) = 2ln13 + 2√13

≈ 12.34

R(12) - R(1) ≈ 12.34-4.21

≈ 8.13

Total reactions to the drugs over the period from t = 1 to t= 12 is approx 8.13.

b) For total reactions from t = 12 to t = 24

When t = 12

R(12) = 2ln13 + 2√13

≈ 12.34

When t = 24

R(24) = 2ln25 + 2√25

≈ 16.44

R(12) - R(1) ≈ 16.44-12.34

≈ 4.10

Total reactions to the drugs over the period from t = 12 to t= 24 is approx 4.10

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3 years ago
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Ksivusya [100]
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5 0
3 years ago
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Can someone help me solve this problem, please!
jok3333 [9.3K]
Answer is 22.25, hope this helps!!
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